Skip to content
Part CCSIR NET December 2024no-mean-and-still-a-weak-law

No mean and still a weak law

Let X1,X2,…,XnX_{1}, X_{2}, \dots, X_{n} be independent and identically distributed (i.i.d.) random variables such that for x=3,4,…,P(X1=±x)=1/(2x = 3, 4, \dots, P(X_{1} = \pm{}x) = 1/(2cx2ln⁡(x))^{2}\ln(x)), where c=∑(x=3..∞)1/(x2ln⁡(x))c = \sum(x=3..\infty) 1/(x^{2}\ln(x)). Then, which of the following statements are true?

  1. A.E(|X1X_{1}|)=∞) = \infty✓ Correct
  2. B.(1/n)∑(i=1..n)Xi→0(1/n)\sum(i=1..n)X_{i} \to 0 in probability as n→∞n \to \infty✓ Correct
  3. C.E(1/|X1X_{1}|)<∞) < \infty✓ Correct
  4. D.E(X12)=∞E(X_{1}^{2}) = \infty✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests moments and tails.

See pricing

50 are analysed free — try those first.

The trap it tests

Moments and tails

A moment assumed to exist, or tail behaviour assumed to be tame.

Drill statements like this

Related counterexample: Convergence in probability implies almost sure convergence

More on this topic

The chapter behind this: Modes of convergence and the limit theorems — free to read

From Limit Theorems and Markov Chains › Modes of convergence, WLLN, SLLN, CLT

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram